Forces of static friction question

[chiuda]

Homework Statement

a 1450 kg car is towing a trailer of mass 454 kg. The force of air resistance is 7471N[backward]. If the acceleration of both vehicles is 0.225m/s^2[forward], what is the force of static friction on the wheels from the ground?

Homework Equations

I am able to find Force normal(Fn) by multiplying its total mass by force of gravity because force normal is equal to force gravity (Fg) because the car is neither hovering or sinking in the surface of which it moves across. I can also find Fnet horizontal (Fnet,h) because i have mass and i have acceleration. but even with these values i am still missing the coefficient for static friction and force applied (Fapp) and force of friction static (Ffs)

The Attempt at a Solution

Fnet,vert=Fn+(-Fg)
0=Fn-Fg
Fn=Mxg
Fn=1904kgx9.81m/s^2
Fn=18678N

Fnet, horiz= MxA
= 1904kgx9.81m/s^2
= 4.4N

I then do not really no what to do other than:

U(coefficient of static friction)

Ffs=UxFn
=Ux18678N

Fnet, horiz=(-F,air)+(-Ffs)+Fapp
4.4N=-7471-Ffs+Fapp
7475.4=-Ffs+Fapp

I am now stuck as you can see and have made many othe attempts in different ways but they all ended up fairly similar, ANY HELP AT ALL WILL BE VERY MUCH APPRECIATED!:)

 

[chiuda]

Sorry just saw that 2. Says relevant equations not questions, I just read it over to quick I guess

 

[chiuda]

Relevant equations are
Ffs=Fn x U
Fnet= M x A

I'm pretty sure that's all

 

[1MileCrash]

The truck is accelerating from a net force.

So what's the net force?
What's the contribution from air resistance?
What's left?

 

[PhanthomJay]

Homework Statement

a 1450 kg car is towing a trailer of mass 454 kg. The force of air resistance is 7471N[backward]. If the acceleration of both vehicles is 0.225m/s^2[forward], what is the force of static friction on the wheels from the ground?

Homework Equations

I am able to find Force normal(Fn) by multiplying its total mass by force of gravity because force normal is equal to force gravity (Fg) because the car is neither hovering or sinking in the surface of which it moves across. I can also find Fnet horizontal (Fnet,h) because i have mass and i have acceleration. but even with these values i am still missing the coefficient for static friction and force applied (Fapp) and force of friction static (Ffs)

The static friction force betweeen the driving wheels and the ground is the applied force which drives the car and trailer forward. You don't need to know the static friction coefficient, only the static friction force.

The Attempt at a Solution

Fnet,vert=Fn+(-Fg)
0=Fn-Fg
Fn=Mxg
Fn=1904kgx9.81m/s^2
Fn=18678N

This is the total normal force acting on the car and trailer upward at the wheels, but you don't need it.

Fnet, horiz= MxA
= 1904kgx9.81m/s^2
= 4.4N

the acceleration is given as 0.225 m/s^2 , not 9.81 m/s^2.

I then do not really no what to do other than:

U(coefficient of static friction)

Ffs=UxFn
=Ux18678N

Fnet, horiz=(-F,air)+(-Ffs)+Fapp
4.4N=-7471-Ffs+Fapp
7475.4=-Ffs+Fapp

I am now stuck as you can see and have made many othe attempts in different ways but they all ended up fairly similar, ANY HELP AT ALL WILL BE VERY MUCH APPRECIATED!:)

Correct your Fnet = ma equation to solve for the correct value of the net force. The net force consists of the friction force and air drag force, and must be in the direction of the acceleration.

 

[chiuda]

The Fnet=MxA has the correct answer I accidentally put in 9.8 m/s^2 instead of 0.225m/s^2. Regardless I has the right value for the answer. You say that Fnet is equal to air drag force and friction force but wouldn't Fnet also include Fapp which I do not have leaving me with two unknowns?

 

[chiuda]

?

 

[chiuda]

I appologize the Fnet is 428N, even so I am still missing the force app and force of friction

 

[1MileCrash]

How is Force applied different from friction in this case? How do you think the truck is applying force?

 

[chiuda]

Wouldn't friction be acting against Fapp?

 

[1MileCrash]

The friction isn't with the truck. It's between the road and tires. The bottom of the tires are moving opposite the truck, making friction in the same direction of the truck.

Think about it, don't you want a lot of friction between the road and your tires? Do you move better or worse on an icy road?

Furthermore, I want you to realize that a frictional coefficient would have been meaningless here because there is nothing to suggest that static friction is at its max value (recall that it is an adjusting force.)

Hope this helps.

 

[chiuda]

Ok I kinda understand

Fnet=Ffs+Fair
428N=Ffs+(-7471N)
428N+7471N=Ffs
Ffs=7899N

Which is the answer in the book but I am just trying to understand it a bit better. If the truck is accelerating wouldn't their have to be an applied force? Is it just because the Fapp is almost equal to the Ff because it has such a low acceleration? I get that the tires are moving backwards in the direction of Ff but doesn't that propell the truck forward?

 

[1MileCrash]

No, the applied force does exist, and its source is the static friction.

Frictional force and applied force aren't "almost equal," they are two names for the exact same thing, here.

 

[1MileCrash]

If your wheels had zero friction with the road you wouldn't accelerate at all. That should tell you that the applied force is friction itself.

That's why wheels are so badass.

 

[chiuda]

Haha good old wheels :P, alright well you say that friction is Fapp HERE could you give me and example of when it wouldn't be? I think that woul help me get the final piece of the puzzle.

 

[1MileCrash]

I think it would be better for you to realize that the term "applied force" is arbitrary. It doesn't really matter what I call the applied force at all.

All you need to know is that acceleration is due to a net force, which is the sum of all forces.

All of your forces here are air resistance and static friction. Thus, they compose the net force and are responsible for acceleration. There is no third, mysterious "applied" force, it is simply a name we give to one of the existing forces.

 

[1MileCrash]

For example, if I reworded the question to say "I'm blowing really hard to try to stop the truck" then I could call air resistance the "applied force" and that really has no effect on the question at all.

 

[chiuda]

Then from what you stated wouldn't the force of friction be causing it to move forward but their would have to be some other friction opposing that motion or is that taken into account for Ffs? And isn't static friction a amount ranging to some maximum unroll the object starts to move, so since their is an acceleration their isn't static but only kinetic friction?

 

[1MileCrash]

The friction opposing motion is actually the air resistance.

As for your other question, it is static friction. There is no kinetic friction involved unless the wheels start slipping (you're peeling out.) If your wheels are not slipping, the wheels are not sliding across the road.

In fact, if there were no acceleration, there is no friction at all, static or kinetic.

 

[chiuda]

K yess I think I just clicked the air was opposing it to move and the static friction was actually opposing this in the other direction making it the "applied" force. K I have one last question I think. Even though the Ffs is acting in the opposite direction wouldn't their be a little bit of friction acting in the same direction as air because the tires are touching the ground or is it because of the Ffs the car is able to move forward and their would only be friction in the same direction as Fair if the brakes were on and the tires were not rotating?

 

[1MileCrash]

Tire friction opposes the tire's direction of motion. There is no tire friction that goes in the same direction of motion of the tires. Friction never does that.

 

[chiuda]

Kk I'm good now thank you so much for the help I really appreciate it I wasn't thinking about the tires direction I was thinking about the car body itself.

 

[PeterO]

Homework Statement

a 1450 kg car is towing a trailer of mass 454 kg. The force of air resistance is 7471N[backward]. If the acceleration of both vehicles is 0.225m/s^2[forward], what is the force of static friction on the wheels from the ground?

Homework Equations

I am able to find Force normal(Fn) by multiplying its total mass by force of gravity because force normal is equal to force gravity (Fg) because the car is neither hovering or sinking in the surface of which it moves across. I can also find Fnet horizontal (Fnet,h) because i have mass and i have acceleration. but even with these values i am still missing the coefficient for static friction and force applied (Fapp) and force of friction static (Ffs)

The Attempt at a Solution

Fnet,vert=Fn+(-Fg)
0=Fn-Fg
Fn=Mxg
Fn=1904kgx9.81m/s^2
Fn=18678N

Fnet, horiz= MxA
= 1904kgx9.81m/s^2
= 4.4N

I then do not really no what to do other than:

U(coefficient of static friction)

Ffs=UxFn
=Ux18678N

Fnet, horiz=(-F,air)+(-Ffs)+Fapp
4.4N=-7471-Ffs+Fapp
7475.4=-Ffs+Fapp

I am now stuck as you can see and have made many othe attempts in different ways but they all ended up fairly similar, ANY HELP AT ALL WILL BE VERY MUCH APPRECIATED!:)

back to square 1.

I think you have forgotten the original question.

The two bits I have highlited red show you the Net force.

F_net = ma

m = 1904 kg
a = 0.225 ms^-2

From that you can calculate the nett force.

Lets suppose the answer to that is 400N (you can calculate the real value)

What makes up the net force?

It is usually the sum of a Forward Force and a backwards force.

The backwards force is the air resistance - given as 7471N

If the net Force is indeed 400N (forward), there must be a forward force of 7841N ( you can again calculate what the real value has to be) acting on the car.

What could be supplying that Force?

Hint: Unless it is due to a gravitational Field [that is down; not forward], magnetic Field (no) or electric Field (no) it must be a CONTACT force.

What is touching the car - or some part of the car?

Hint 2: It is not the air touching the car - that has already been taken care of in the air resistance figure of 7471N [backward].

EDIT: Whoops, while I was composing I think you were finding the solution.